
Two free positive charges \[4q\] & \[q\] are at a distance \[l\] apart. What charge \[Q\] is needed to achieve equilibrium for the entire system & where should it be placed to form charge \[q\]?
Answer
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Hint: In order to solve this question, we are first going to find the total force on the two charges \[4q\] & \[q\] due to the equilibrium charge \[Q\]. Then, the distance of the charge is found by applying the equilibrium condition. After that the value of the equilibrium charge \[Q\] is found by further solving.
Formula used:
The total force on the two charges \[4q\] & \[q\] at a distance \[l\]apart is given by
\[F = \dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{\left( {l - r} \right)}^2}}}\]
Complete step-by-step solution:
Let us consider that the charge \[Q\] is in equilibrium and is present at a distance \[r\] from the charge \[q\].
The total force acting on the charges \[4q\] and \[q\] is given by
\[F = \dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{\left( {l - r} \right)}^2}}}\]
For the charge\[Q\]to be in equilibrium, the total force has to be equal to zero.
\[\dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{\left( {l - r} \right)}^2}}} = 0\]
Solving this equation, we get
$ \Rightarrow {\left( {l - r} \right)^2} = 4{r^2} \\
\Rightarrow \left( {l - r} \right) = 2r \\
\Rightarrow l = 3r \\
\Rightarrow r = \dfrac{l}{3} \\ $
Now, taking the third charge to be present there, and considering the charge to be equal to \[ - Q\], and on applying the equilibrium condition on the charge \[ + q\],
$ \dfrac{{kQ}}{{{{\left( {\dfrac{l}{3}} \right)}^2}}} = \dfrac{{k4q}}{{{l^2}}} \\
\Rightarrow Q = \dfrac{{4q}}{9} \\ $
Therefore a point charge \[ - \dfrac{{4q}}{9}\] should be placed at a distance of \[\left( {\dfrac{l}{3}} \right)\] rightwards from the point charge\[ + q\] on the line joining the two charges.
Note: It is important to note that the total force on the two charges when they are present in equilibrium with each other is equal to zero. This fact has been used twice in this question in order to find the distance of the charge in equilibrium and the magnitude of that charge also.
Formula used:
The total force on the two charges \[4q\] & \[q\] at a distance \[l\]apart is given by
\[F = \dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{\left( {l - r} \right)}^2}}}\]
Complete step-by-step solution:
Let us consider that the charge \[Q\] is in equilibrium and is present at a distance \[r\] from the charge \[q\].
The total force acting on the charges \[4q\] and \[q\] is given by
\[F = \dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{\left( {l - r} \right)}^2}}}\]
For the charge\[Q\]to be in equilibrium, the total force has to be equal to zero.
\[\dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{\left( {l - r} \right)}^2}}} = 0\]
Solving this equation, we get
$ \Rightarrow {\left( {l - r} \right)^2} = 4{r^2} \\
\Rightarrow \left( {l - r} \right) = 2r \\
\Rightarrow l = 3r \\
\Rightarrow r = \dfrac{l}{3} \\ $
Now, taking the third charge to be present there, and considering the charge to be equal to \[ - Q\], and on applying the equilibrium condition on the charge \[ + q\],
$ \dfrac{{kQ}}{{{{\left( {\dfrac{l}{3}} \right)}^2}}} = \dfrac{{k4q}}{{{l^2}}} \\
\Rightarrow Q = \dfrac{{4q}}{9} \\ $
Therefore a point charge \[ - \dfrac{{4q}}{9}\] should be placed at a distance of \[\left( {\dfrac{l}{3}} \right)\] rightwards from the point charge\[ + q\] on the line joining the two charges.
Note: It is important to note that the total force on the two charges when they are present in equilibrium with each other is equal to zero. This fact has been used twice in this question in order to find the distance of the charge in equilibrium and the magnitude of that charge also.
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